Question

(ii) Mole fraction of 1.8 g glucose in 90 g water is :
(a) 0.19
(b)0.019
(c)0.0019
Lii Molarity in mol/l of distilled or pure water is.​

Answer 1

Answers:-

1) Given:

Mass of glucose = 1.8 g

Formula of glucose → C₆H₁₂O₆

• Molar mass of glucose = 6(12) + 12(1) + 6(16) = 72 + 12 + 96 = 180 g/mol

Mass of water = 90 g

Formula of water → H₂O

• Molar mass of water = 2(1) + 1(16) = 2 + 16 = 18 g/mol.

We know that,

number of moles (n) = Mass in grams / molar mass

So,

★ n of glucose = 1.8/180 = 0.01 moles.

★ n of water = 90/18 = 5 moles.

Now,

Mole fraction of a substance = No.of moles of the substance / Total no.of moles of the solution

Simply,

Xₓ = nₓ / nₓ + nᵧ

We have to find the mole fraction of glucose.

So,

• Number of moles of glucose = 0.01 mol

• Total number of moles in the solution = 0.01 + 5 = 5.001 mol

Hence,

⟶ Mole fraction of glucose = 0.01/5.001 = 0.0019

∴ The mole fraction of glucose is 0.0019 (Option - C)

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2)

we have to find molarity of pure/distilled water (in mol/l).

We know that,

Molarity (M) = Number of moles of substance / Volume of the solution

We already have volume of solution i.e., 1 l.

We know,

Number of moles = Mass /Molar mass.

• Molar mass of water = 18 g/mol

Mass can be defined as product of volume and density.

• Density of pure water is 0.99 g/ml or = 1 g/ml

So,

Mass of water = 1 g/ml × 1 l

• 1 litre = 10³ ml

⟶ mass of water = 1 g/ml × 10³ ml

⟶ mass of water = 10³ g

Now

⟶ n of water = 10³/18

⟶ n of water = 1000/18

⟶ n of water = 55.55

Hence,

Molarity of pure water = 55.55 / 1 = 55.55 M

∴ The molarity of pure water is 55.55 M.

Answer 2

Given: Given mass of glucose = 1.8 grams

Given mass of water = 90 grams

To Find: (i) Mole fraction of the solution

(ii) Molarity of water

Solution:

(i)  To determine the mole fraction of the solution, first determine the number of moles of each components i.e both solute and the solvent.

• The number of moles can be determined by dividing the given mass of the substance to the molar mass of the substance.
• Firstly, calculate the number of moles of glucose
• Formula of glucose is $C_{6} H_{12} O_{6}$. So, the molar mass of the glucose is

           = (6 × 12) + (12×1) + (6×16) = 72 + 12 + 96 = 180 g/mol.

• The number of moles of glucose = $\frac{1.8}{180}$  = 0.01 moles.
• Formula of water is $H_{2}O$. So, the molar mass of the water is

        = (1×2) + 16 = 18 g/mol.

• The number of moles of water = $\frac{90}{18}$ = 5 moles.
• Now, the formula of mole fraction of a substance is equals to the number of moles of the glucose divided by the total number of moles of the solution.

        = $X_{x} = \frac{n_{x} }{n_{x} + n_{y} }$  = $X_{x} = \frac{0.01}{0.01 + 5}$

         = $\frac{0.01}{5.01}$ = 0.0019 moles.

(ii) To determine the molarity of the water the following formula is used.

• Molarity (M) = number of moles of water is divided by the volume of the solution.

• Volume of the solution = $\frac{mass}{density}$
• Density of pure water = 1g/ml
• As we know that the volume of the solution is 1 litre = 1000 ml
• Mass of water = volume × density
• 1000 ml = 1000 grams
• mass of water = 1g/ml × 1000 g
• Now, number of moles of water = $\frac{1000}{18}$ = 55.55 moles
• Molarity (M) = $\frac{55.55}{1}$ = 55.55 mol/l.

Hence, the mole fraction of glucose is 0.0019 and the molarity of water is 55.55 mol/l.